What is the integral of (8x^3+1/(x^2)) ?

The integral of (8x^3+1/(x^2)) is 2x^4-1/x+C

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integral of (8x^3+1/(x^2))

\int (8 x^{3} + \frac{1}{x ^{2}}) d x

2 x^{4} - \frac{1}{x} + C

Solution steps

\int 8 x^{3} + \frac{1}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 8 x^{3} d x + \int \frac{1}{x ^{2}} d x

\int 8 x^{3} d x = 2 x^{4}

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= 2 x^{4} - \frac{1}{x}

Add a constant to the solution

= 2 x^{4} - \frac{1}{x} + C

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What is the integral of (8x^3+1/(x^2)) ?
The integral of (8x^3+1/(x^2)) is 2x^4-1/x+C